# Show that $\tfrac{1}{2}(b B\cdot (c \wedge a) - a B\cdot (c \wedge b)) = (c\cdot b) a \cdot B$ in geometric algebra

I would like to show that

$$\frac{1}{2}\left( b B\cdot (c \wedge a) - a B\cdot (c \wedge b)\right) = (c\cdot b) a \cdot B$$

where B is a bivector, and the others are vectors. I've tried to coerce every term into having a factor of $$a\cdot B$$, but this just introduces other terms.

Edit: some background: Hestenes claims

\begin{aligned} G^\beta &=\tfrac{1}{2} (g^\beta \wedge g^\mu \wedge g^\nu) \cdot \omega_{\mu\nu} \\ &= g^\mu \cdot \omega_{\mu\nu} g^{\nu \beta} -\tfrac{1}{2}g^\beta (g^\nu \wedge g^\mu)\cdot \omega_{\mu\nu} \end{aligned}

Apply Equ (1.40) (Hestenes & Sobczyk 1984) \begin{aligned} B_r \cdot (a_1 \wedge \cdots \wedge a_n) = \sum_{j_1 < \cdots < j_r} \epsilon(j_1, \dots, j_n) B_r \cdot (a_{j_1} \wedge \cdots \wedge a_{j_r}) a_{j_{r+1}} \cdots a_{j_n} \end{aligned}

Switch the order of the inner product (gives factor of +1) and then apply the formula,

\begin{aligned} G^\beta = \tfrac{1}{2} \left( g^\nu \omega_{\mu\nu} \cdot (g^\beta \wedge g^\mu) - g^\mu \omega_{\mu\nu} \cdot (g^\beta \wedge g^\nu) + g^\beta \omega_{\mu\nu} \cdot (g^\mu \wedge g^\nu) \right) \end{aligned}

The last term is equal to $$-\tfrac{1}{2}g^\beta (g^\nu \wedge g^\mu) \cdot \omega_{\mu\nu}$$; the remainer is the problem in question, with $$a=g^\mu$$, $$b=g^\nu$$, $$c=g^\beta$$, $$B=\omega_{\mu\nu}$$.

• Your subject and body equations don't match. Which one is the one you are trying to show? Nov 11, 2023 at 1:44
• Where is this coming from? You can turn the LHS (without the $1/2$) into $(B\cdot c)\cdot(a\wedge b)$, but the RHS doesn't make sense to me. Nov 11, 2023 at 5:23
• I added a factor 2 into the title and a couple cosmetic changes Nov 11, 2023 at 5:52
• This is a subproblem of showing that equ 148 is the same as the first part of equ 146. $a = g^\mu$, $b = g^\nu$, $c = g^\beta$, of researchgate.net/publication/… Nov 11, 2023 at 6:01
• I think you are looking at this wrong and that the identity you've written is false. It seems to me that the derivation you want from Eq. 146 to 148 involves index renaming, the fact that $\omega_{\mu\nu}$ is antisymmetric, and the fact that $g^{\mu\nu} = g^\mu\cdot g^\nu$ is symmetric. Nov 11, 2023 at 16:47

The generalized identity you present is almost certainly false, and instead the derivation you seek relies on the specific problem at hand.

Begin with $$G^\beta = \frac12(g^\beta\wedge\omega_{\mu\nu})\cdot(g^\mu\wedge g^\nu)$$ as Hestenes does. Then apply the two identities he suggests successively to get $$G^\beta = \frac12(g^\beta\wedge\omega_{\mu\nu})\cdot(g^\mu\wedge g^\nu) = \frac12[g^\beta\wedge(\omega_{\mu\nu}\cdot g^\mu) + g^{\beta\mu}\omega_{\mu\nu}]\cdot g^\nu$$$$= \frac12[g^\beta(\omega_{\mu\nu}\cdot g^\mu)\cdot g^\nu - g^{\beta\nu}\omega_{\mu\nu}\cdot g^\mu + g^{\beta\mu}\omega_{\mu\nu}\cdot g^\nu]. \tag{*}$$

The first term of ($$*$$) becomes $$g^\beta(\omega_{\mu\nu}\cdot g^\mu)\cdot g^\nu = g^\beta\omega_{\mu\nu}\cdot(g^\mu\wedge g^\nu) = -g^\beta(g^\nu\wedge g^\mu)\cdot\omega_{\mu\nu}$$ using the commutivity of the dot product and anticommutivity of the wedge product in this case.

The second term becomes $$-g^{\beta\nu}\omega_{\mu\nu}\cdot g^\mu = g^\mu\cdot\omega_{\mu\nu}g^{\beta\nu}$$ using the anticommutivity of the dot product in this case and the fact that $$g^{\beta\nu}$$ is a scalar.

For the third term of ($$*$$) we do the same thing but then relabel indices and use the antisymmetry of $$\omega_{\mu\nu}$$ to get $$g^{\beta\mu}\omega_{\mu\nu}\cdot g^\nu = -g^\nu\cdot\omega_{\mu\nu}g^{\beta\mu} = -g^\mu\cdot\omega_{\nu\mu}g^{\beta\nu} = g^\mu\cdot\omega_{\mu\nu}g^{\beta\nu}.$$

Combining all the terms now yields $$G^\beta = -\frac12g^\beta(g^\nu\wedge g^\mu)\cdot\omega_{\mu\nu} + g^\mu\cdot\omega_{\mu\nu}g^{\beta\nu}$$ as desired.

I do not know how Hestenes concluded that $$G^\beta = \frac12(g^\beta\wedge g^\mu\wedge g^\nu)\cdot\omega_{\mu\nu} = \frac12(g^\beta\wedge\omega_{\mu\nu})\cdot(g^\mu\wedge g^\nu)$$ nor how his Eq. (147) could at all be relevant. What we can derive easily is $$\frac12(g^\beta\wedge g^\mu\wedge g^\nu)\cdot\omega_{\mu\nu} = -\frac12(g^\beta\cdot\omega_{\mu\nu})\cdot(g^\mu\wedge g^\nu)$$ which makes me feel like the previous equation is incorrect, but I have no proof of this.

The quantity

\begin{aligned} \tfrac{1}{2} (c \wedge a \wedge b) \cdot B \end{aligned}

must contain a projection between $$B$$ and $$c$$, but the following does not:

\begin{aligned} a \cdot B (b\cdot c) - \tfrac{1}{2}c (b \wedge a) \cdot B \end{aligned}

Therefore they cannot be equal, contrary to what Hestenes claims. Therefore the subproblem statement given in the question cannot be correct either.

• This is in no way a rigorous argument. It also does not imply that Hestenes' identity is false since the quantities involved are not generic and instead have properties beyond just being vectors or bivectors. As my answer shows you can derive Eq. (148) by starting from the RHS of Eq. (146), but I do note that perhaps the second equality in Eq. (146) is incorrect or at least requires more justification than Hestenes provides. Nov 13, 2023 at 0:00
• Regardless of any other properties they have, they are vectors and bivectors, so the equality fails for the reason given in this answer Nov 13, 2023 at 4:38
• "Therefore they cannot be equal, contrary to what Hestenes claims" is what you said; but Hestenes made no claim about your generic equality, and whether or not your generic equality is true has no bearing on Hestenes' claim. And again, you have made a qualitative observation which suggests that your identity is false, but it does not in any way constitute a proof that it is false. Nov 13, 2023 at 13:55
• The idea that omega has no projection with g strains plausibility, but you can post a proof if you find it Nov 13, 2023 at 16:28